Puzzles with numbers and things

[abextra]

R. 349: multiplying balls
Those 4 apes like playing around with their shiny snooker balls. This time they play a game of snookwhiz, each ape starts out with 7 reds and additional balls can be won over the duration of the game. In the end they all add up their point scores according to the balls in hand. Turns out they have between 7 and 23 points.
Charlie ponders those numbers a bit and then thinks about asking you guys this:
Say we’d multiply each ape’s score by each of the other 3 apes’ scores and then multiply those 3 results, then get even crazier and take the product of the 4 results. Now divide that by the product of each of our scores and at that point discard any fractional portion. Now we simply subtract the product of our point scores divide the resulting value by 10, and the question is:

“What is the remainder?”

(Please add a short explanation, of course)

Hi, Snookersfun!

What does Charlie mean by " discard any fractional portion"? Should I try to spell RED HERRING?

[snookersfun]

breaking news!

moglet first in balancing (or rather tilting). Congratulations!

[snookersfun]

and the promised

R 350: tilted mobile
tilted.bmp

this time 3 sets of reds all distributed into the bags on the mobile (all different amounts as before). Each balance is tilted one unit to the right (or total torque on right side is one unit greater than on left)

[snookersfun]

thanks for all the solutions moglet and abextra

and I can report: tallguy first in on R. 349 and with a proper sophisticated explanation as well

and moglet and Mon right on tallguy's feet! Congrats as well

[abextra]

Let's also close those pentaminoes, here are the pictures I received from Mon and moglet for that round (R. 344). If anybody should have the wish to have a go at those, there are plenty more unique solutions (1010 all together)
well done, moglet!

well done, Mon

and well done Abextra, who's solution I can't find

Would have been very suspicious if you could find my pic - as it was in my PC only till now (cos I'm an idiot at computers and couldn't make it to move forward) - but here it is.

Wasn't easy for me to get the solution, hard to believe there are 1010 arrangements and only one of them revealed itself to me!

[moglet]

340

[ATTACH]1851[/ATTACH]

346, shortly..

[snookersfun]

moving on

Alright, to pick up now (due to popular demand -> abextra) after the Christmas slump, but before going into the New Year’s slump:

There will be a tilted mobile later on (too much spiked eggnog),… if I can dig it up at home

But to ease in, for now
R. 349: multiplying balls
Those 4 apes like playing around with their shiny snooker balls. This time they play a game of snookwhiz, each ape starts out with 7 reds and additional balls can be won over the duration of the game. In the end they all add up their point scores according to the balls in hand. Turns out they have between 7 and 23 points.
Charlie ponders those numbers a bit and then thinks about asking you guys this:
Say we’d multiply each ape’s score by each of the other 3 apes’ scores and then multiply those 3 results, then get even crazier and take the product of the 4 results. Now divide that by the product of each of our scores and at that point discard any fractional portion. Now we simply subtract the product of our point scores divide the resulting value by 10, and the question is:

“What is the remainder?”

(Please add a short explanation, of course)

[snookersfun]

thank you, tallguy!

Let's also close those pentaminos, here are the pictures I received from Mon and moglet for that round (R. 344). If anybody should have the wish to have a go at those, there are plenty more unique solutions (1010 all together)

well done, moglet!

pentaminos12x5Sol.jpg well done, Mon

and well done Abextra, whose solution I can't find

and if anybody can put up the solutions to those balance rounds R.338, 340 and 346, we could close those as well.

[tallguy]

Solution to R.347 is

-- 10 05 -- -- --
12 -- -- -- 08 13
-- 07 01 04 06 --
-- -- 02 -- 11 --
-- -- 09 03 -- --

The distances b/w pairs.

Pair Distance squared
13,12 25
12,11 20
11,10 18
10,9 17
9,8 13
8,7 10
7,6 9
6,5 8
5,4 5
4,3 4
3,2 2
2,1 1


How I arrived at the sol.

Consider the grid to have origin at top left and adjacent parallel lines to be unit distance apart. Horizontal axis is X axis and vertical is Y axis.

After a few minutes of analysis I figured out that there weren't too many unique distances that were possible since balls have to placed at (x,y) where x and y are both integers and x<=5,y<=4.

There turned out to be 13 unique distances as in the following table.

coordinate diff - distance squared
(note / means OR)

(5,0) - 25
(4,2) / (2,4) - 20
(4,1) / (1,4) - 17
(4,0) / (0,4) - 16
(3,3) - 18
(3,2) / (2,3) - 13
(3,1) / (1,3) - 10
(3,0) / (0,3) - 9
(2,2) - 8
(2,1) / (1,2) - 5
(2,0) / (0,2) - 4
(1,1) - 2
(1,0) / (0,1) - 1


There are 12 distances b/w pairs.Since the distance b/w a pair (k+1,k) is strictly greater than the previous pair (k,K-1) the 12 distances will have be to selected from these 13 unique distances. In other words 1 out of the 13 distances will not be present. This leaves us with 13 possibilites.

I couldn't narrow it down any further. So I began trying all of them out using a backtracking method.Started with (13,12) as (5,0). Luckily I didn't encounter multiple positional choices for a distance and therefore had to backtrack only once ( at the point where a distance was skipped).

[snookersfun]

...and also a late entrant to R.347 now. Tallguy solved that one, well done!
If he could put up his solution (and the thorough explanation) we could close that round at least.

again congrats to all entrants!

[abextra]

Good news, thank you, ladies.

Is anyone else working on these questions? Quite many opened rounds now - would be nice to see some of the answers and then maybe even move on... (dreaming, lol).

[Monique]

and 342/2 aswell ... well done!

and belated but nevertheless sincere apologies to Kathrin for the round numbers collision. I did not mean to ignore your contribution Kathrin.

[Kathrin]

d_g also solved R342/1

sorry it took me a while to post it

[snookersfun]

Hi d_g, unless I miss my hunch the distances puzzle still holds good in 3 dimensions, the mean height of one stack measured to the mean height of the subsequent stack and so on......

...actually that sounds plausible, if one would stack strictlty vertically...
but alright, I admit, stack wasn't the word to use...maybe

and of course, that means that also moglet had no problem in solving it
well done! :snooker:

[moglet]

Hi d_g, unless I miss my hunch the distances puzzle still holds good in 3 dimensions, the mean height of one stack measured to the mean height of the subsequent stack and so on......